According to the Bohr model of a hydrogen atom, an electron revolves around the nucleus in circular orbits. What is the ratio of the magnetic dipole moment of the electron in the ground state ( n=1 ) to its magnetic dipole moment in the first excited state ( n=2 )?

Step-by-step solution

The magnetic dipole moment of an electron revolving in the n^ th orbit of a hydrogen atom is given by M = evr 2 . In the Bohr model, the velocity of the electron v 1 n and the radius of the orbit r n^2 . Therefore, the magnetic moment M ( 1 n )(n^2) = n . Alternatively, the magnetic moment is an integral multiple of the Bohr magneton: M = n ( eh 4 m_e ) , which directly shows M n . For the ground state ( n=1 ) and the first excited state ( n=2 ), the ratio of their magnetic dipole moments is M₁ M₂ = 1 2 , or 1:2 .